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Math Help - A basic function limit question

  1. #1
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    A basic function limit question

    We've just started doing continuity, limits of functions etc, and I'm not quite sure how to progress.

    If we suppose g and h are real-valued functions, defined on some interval (m,n) containing <br />
x_0 <br />
, and <br />
\mathop {\lim }\limits_{x \to x_0 } g\left( x \right) = k<br />
and <br />
\mathop {\lim }\limits_{x \to x_0 } h\left( x \right) = l<br />
.

    I want to show that if <br />
g\left( x \right) < h\left( x \right){\text{ }}\forall x \in \left( {x_0  - \delta ,x_0  + \delta } \right),\delta  > 0<br />
then <br />
k \leqslant l<br />

    Also, do you know of any good resources to get further help on this topic? I know the members here are extremely knowledgeable, but do you have any online text resource etc that may help? Maybe with examples of a similar kind?
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  2. #2
    MHF Contributor

    Joined
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    Quote Originally Posted by StandardToaster View Post
    We've just started doing continuity, limits of functions etc, and I'm not quite sure how to progress.

    If we suppose g and h are real-valued functions, defined on some interval (m,n) containing <br />
x_0 <br />
, and <br />
\mathop {\lim }\limits_{x \to x_0 } g\left( x \right) = k<br />
and <br />
\mathop {\lim }\limits_{x \to x_0 } h\left( x \right) = l<br />
.

    I want to show that if <br />
g\left( x \right) < h\left( x \right){\text{ }}\forall x \in \left( {x_0 - \delta ,x_0 + \delta } \right),\delta > 0<br />
then <br />
k \leqslant l<br />
    suppose k > l and choose \epsilon = \frac{k-l}{2}. then by the definition of limit, there exists \delta_1 > 0 such that g(x) > k - \epsilon if |x-x_0| < \delta_1. also there exists \delta_2 > 0 such that h(x) < l + \epsilon if |x-x_0| < \delta_2.

    now let \delta_0=\min \{\delta, \delta_1, \delta_2\}. so if |x-x_0| < \delta_0, then we'll have: l + \epsilon > h(x) > g(x) > k- \epsilon, and thus: \epsilon > \frac{k-l}{2}. contradiction!
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